document.write( "Question 313284: 62(200
\n" ); document.write( "Solution A is 80% Alcohol and Solution B is 30% Alcohol. How many Liters of each Solutions should be mixed in order to make 200 liters of a Solution that is 62% Alcohol ?
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Algebra.Com's Answer #224027 by Fombitz(32388)\"\" \"About 
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1.\"80A%2B30B=62%28200%29\"
\n" ); document.write( "2.\"A%2BB=200\"
\n" ); document.write( "Multiply eq.2 by \"-30\" and add the equations together.
\n" ); document.write( "\"80A%2B30B-30A-30B=62%28200%29-30%28200%29\"
\n" ); document.write( "\"50A=32%28200%29\"
\n" ); document.write( "\"A=32%284%29\"
\n" ); document.write( "\"highlight%28+A-128%29\"
\n" ); document.write( "Then from eq. 2,
\n" ); document.write( "\"128%2BB=200\"
\n" ); document.write( "\"highlight%28+B=72%29\"
\n" ); document.write( "128 liters of 80% solution and 72% of the 30% solution.
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